How Does Electric Flux Change with Different Closed Surfaces Around Dipoles?

[upthedown]

I have a figure showing two dipoles each having a Q and -Q charge with distance d separating the positive and negative of each dipole. The dipoles are then surrounded by multiple closed surfaces. I need to match fluxes of 8pikQ, 4pikQ, -8pikQ, - 4pikQ and 0 to these surfaces.


Gauss' Law states that flux=Eda=4(pi)kq for any closed surface


Since I have dipoles, my electric fields will be pulled toward the negative charge. THerefore, surfaces with negative charges in them should have a negative flux b/c the field points to the interior of the closed surface while my surfaces encompassing positive charges should have positive flux b/c the field points to the exterior of the surface. So, since each surface has a flux of 4pikq wouldn't the flux just keep adding up as the vector went through each additional surface?

thanks

 

[Mindscrape]

Could you post the figure?

 

[m2003]

for your question! You are correct in your understanding that the electric flux would add up as it passes through each additional surface. However, it is important to note that the flux through a closed surface is equal to the net charge enclosed by that surface divided by the permittivity of the medium (in this case, the vacuum). In other words, the total flux through a closed surface is directly proportional to the total charge enclosed by that surface.

In the given scenario, we have two dipoles with equal and opposite charges, so the net charge enclosed by each closed surface will be 0. This means that the flux through each surface will also be 0.

To match the given flux values of 8pikQ, 4pikQ, -8pikQ, -4pikQ, and 0, we would need to have additional charges placed within the closed surfaces. For example, to have a flux of 4pikQ, we could add a positive charge of Q at the center of one of the dipoles. This would create a net charge of Q enclosed by the closed surface, resulting in a flux of 4pikQ. Similarly, to have a flux of -8pikQ, we could add a negative charge of 2Q at the center of one of the dipoles.

In summary, the flux through a closed surface is determined by the net charge enclosed by that surface. In the case of dipoles, the net charge is 0, so additional charges would need to be added to match the given flux values. I hope this helps clarify the concept of electric flux and dipoles.